I’m honestly stuck. I’ve seen people say 1/2, others 2/3. Something Monty Hall Problem, Bayes Theorem but I’m still confused so here I am.
Edit: I believe you are not allowed to change your box choice on the 2nd “turn” as that would make having two boxes pointless, wouldn’t it?
In: 10
Before I explain the solution to the problem, let’s see a different problem with dice, which are easier to understand.
You’ve got a regular six-faced fair die numbered from 1 to 6. What is the probability of getting a 3? When you want to calculate a probability of equally-likely outcomes, there’s a formula, which goes:
* Probability = **Number of desired outcomes** ÷ **Number of possible outcomes**.
When rolling a die, there are **six** possible outcomes, but getting a 3, the desired outcome, is only **one** of them. In this case, the probability would be **1/6**. Similarly, what is the probability of getting either a 4 or a 6? The number of possible outcomes is still **six**, but now there are **two** desired outcomes, so this probability is **2/6** (or 1/3).
What about this: “After rolling a die, the result is *at least 5*. What is the probability that the result was a 6?” We can still use the formula, but we have to be careful. What is the number of possible outcomes? If you say **six**, that’s incorrect, because **we know** that *at least 5*, which means that the die was either a 5 or a 6. Those are the only possible outcomes if the result is at least five, so the number of possible outcomes in this case is **two**. What about the number of desired outcomes? Well, the desired outcome was getting a 6, so only **one** desired outcome. The probability is, then, **1/2**. And it makes perfect sense: if you roll a die and get at least a 5, you either got a 5 or a 6, each equally likely with a 1/2 probability.
Let’s go back to the original problem, which I will restate to make it more clear:
> There are two boxes, box A and box B. In box A there is one $1 bill and one $100 bill. In box B there are two $100 bills.
> You now carry the following experiment: first, choose a box at random, and then pick a bill from said box. Suppose that the result of this experiment is that the bill that you got is a $100 bill. What is the probability that the chosen box was box B?
I changed a bit the problem to make it easier: you asked “what is the chance of taking out another $100 bill”, but that is the same as asking what is the probability that the box that you chose was box B.
What are the possible outcomes of the experiment (without taking into account that we know the result)? Let’s enumerate them:
1. You choose box A and pick the $1 bill.
2. You choose box A and pick the $100 bill.
3. You choose box B and pick one of the $100 bill.
4. You choose box B and pick the other $100 bill.
Each of these outcomes are equally likely (like rolling a die). If I asked: what is the probability of picking a $100 bill? Then, the answer would be: there are **four** possible outcomes, of which **three** of them are desirable, so the probability is **3/4**.
The problem asks: “the picked bill is a $100 bill. What is the probability that the chosen box was box B?” What are the possible outcomes? Even if our experiment has **four** possible outcomes (rolling a die has six outcomes), if we know that we picked a $100 bill, only **three** of them are possible (when rolling a die, *at least five* means two possible outcomes ). Out of those three, how many are desirable? One of them comes from choosing box A, and two of them from choosing box B. Thus, the number of desirable outcomes is **two**. The probability is **2/3**.
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